Chapter 6.4, Problem 24PPS

To determine

To solve the given system of equations using elimination method

Answer to Problem 24PPS

  $x=\frac{1}{2}$and $y=4$

Explanation of Solution

Given:

Equations:

  $\frac{2}{5}x+6y=24\frac{1}{5}\Rightarrow 2x+30y=121$

  $3x+\frac{1}{2}y=3\frac{1}{2}\Rightarrow 6x+y=7$

Calculation:

Simplifying the given equations,

  $\begin{array}{l}\Rightarrow \frac{2}{5}x+6y=24\frac{1}{5}\\ \Rightarrow \frac{2}{5}x+6y=\frac{24\times 5+1}{5}\\ \Rightarrow \frac{2}{5}x+6y=\frac{120+1}{5}\\ \Rightarrow \frac{2}{5}x+6y=\frac{121}{5}\\ \Rightarrow \frac{2x+30y}{5}=\frac{121}{5}[\text{Taking}\text{Least}\text{Common}\text{factor}]\\ \Rightarrow 2x+30y=121\end{array}$

  $\begin{array}{l}\Rightarrow 3x+\frac{1}{2}y=3\frac{1}{2}\\ \Rightarrow 3x+\frac{1}{2}y=\frac{3\times 2+1}{2}\\ \Rightarrow 3x+\frac{1}{2}y=\frac{6+1}{2}\\ \Rightarrow 3x+\frac{1}{2}y=\frac{7}{2}\\ \Rightarrow \frac{6x+y}{2}=\frac{7}{2}[\text{Taking}\text{Least}\text{Common}\text{factor}]\\ \Rightarrow 6x+y=7\end{array}$

Let,

  $\begin{array}{l}2x+30y=\mathrm{121.......}(i)\\ 6x+y=\mathrm{7.......}(ii)\end{array}$

Now, subtract equation $30\times$ (ii) from equation (i),

  $\begin{array}{l}\Rightarrow (2x+30y)-30\times (6x+y)=121-30\times 7\\ \Rightarrow 2x+30y-180x-30y=121-210\\ \Rightarrow -178x=-89\\ \Rightarrow x=\frac{1}{2}\end{array}$

Now, substitute the value of x in equation (i),

  $\begin{array}{l}\Rightarrow 2\times \frac{1}{2}+30y=121\\ \Rightarrow y=4\end{array}$

Conclusion:

Therefore, the required solution for the given system of equation is $x=\frac{1}{2}$and $y=4$ .